17. Letter Combinations of a Phone Number

description:

Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.

telephone buttons【拼音输入法九键拿自己手机看一眼】

Example:

Input: "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

my answer:

感恩

my answer

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        unordered_map<int, char>m = {'1':{''}, 
                                     '2':{'a','b','c'},
                                     '3':{'d','e','f'},
                                     '4':{'g','h','i'},
                                     '5':{'j','k','l'},
                                     '6':{'m','n','o'},
                                     '7':{'p','q','r','s'},
                                     '8':{'t','u','v'},
                                     '9':{'w','x','y','z'},
                                     '0':{''}};
        string res = '';
        vector<vector<int>> ready;
        for (int i = 0; i < digits.size(); ++i){
            ready.push_back(m[digits[i]]);
            not finished...............from begin to give up
            ##############################################################
            # 参考大佬的coding后恍然大悟，为什么我没有想到递归呐，因为就是
            # 类似于深度优先遍历这种，从一个点扎下去然后不断地扩散，典型递
            # 归的使用                                                 
            ##############################################################
        }
    }
};

大佬的answer：

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        if (digits.empty()) return {}; //return的vector是空不是return "";是return {};
        vector<string> res;
        string dict[] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        letterCombinationsDFS(digits, dict, 0, "", res);//这里传进去的是dict, not dict[]
        return res;
    }
    void letterCombinationsDFS(string digits, string dict[], int level, string out, vector<string> &res) {
    //这里用&res是说这个东西是引用，就是里边对res的操作会对外边的res起作用，所以返回值才是void，要不然就是vector<string>了
        if (level == digits.size()) {res.push_back(out); return;}
        string str = dict[digits[level] - '0'];
        for (int i = 0; i < str.size(); ++i) {
            letterCombinationsDFS(digits, dict, level + 1, out + string(1, str[i]), res);
        }
    }
};

relative point get√：

• string s(num,c) //生成一个字符串，包含num个c字符
• for (auto x : iterable something)
• 静态数组 int array[100] = {1，2};

hint :